Hello!! Is anyone good with evaluating indefinite integrals?

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Hello!! Is anyone good with evaluating indefinite integrals?

Mathematics
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This one seems easy but I could sure use confirmation on the correct answer!! Lol.
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Posting!!

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Other answers:

uh uhs, amistre is here :P
.......TRIG......
i meant uh ohs*
I couldn't figure out how to post a division with out the sign!
give me a T; give me an R; give me an IG...whats that spell? trouble?
lols nah, spells awesomeness
post it your best and well decipher it :)
\frac{numer}{deno}
ArcsinX + C
^-- in the equation editor and like amistre said it's not a big deal
lols all yours amistre
integrate arcsin + C? or is that your answer that yougot?
That's my answer - I just have to evaluate the integral
whats the integral so that we can pervue its intricate innards :)
\[\frac{1}{\sqrt{1 - x^2}}\] is this the integral?
without the integral sign...or the dx...
\frac\[\int\limits_{?}^{?}\frac{1}{1+x^2} dx\]
thats a arctan..
yup, amistre's right
i saw it in a movie lol
lols "down the rabbit hole"?
:) tan^1(x) + C...are there any bounds to evaluate?
..^-1..
\[[\int\limits] dx 1 + x2 = \arctan x + C \]
Was that all I needed to refer to?! I have it in my notes but the 'dx' was in the num
its like learning greek all over again ;)
something like this? \[\int\limits_{}^{}\frac{1}{x^2 + 1}dx\]
Noooo kidding! lol
Noooo kidding! lol
\[\int\limits_{} \frac{1}{1+x^2} dx \rightarrow \tan^{-1}(x) + C\]
for answering this question, amistre...i give you...a MEDAL
-applause-
yay!! ..... its prolly foiled covered chocolate...so I accept :)
lols how'd you guess? enjoy! cheers
Will arctan x + c be correct?
-nod-
as long as you havent left anyting out, yes; but arctan is a dated term
i would know; im dated ;)
lol!
Can you guys help me with another problem/confirm my answer being right or wrong?!!
perhaps, but display the question first; just makes it easier :)
lol - kk!!
Eval the Indefinite Integral :\[\int\limits\limits_{?}^{?}\frac{x^3 dx}{\sqrt[4]{5x^4-1}}\]
you want a 15 for that :)
I got: 1/15 (5x^4)^3/4 + C
its refered to as "u- substitution" u = 5x^4 - 1; du = 20x^3 dx dx = du/20x^3
\[\frac{1}{15} (5x^{4})^{3/4} + C\]
\[\int\limits_{} \frac{x^3 }{20x^3 u^{1/4}}du\]
i mistyped 15; meant 20 :) the x^3s cancel; pull out the 1/20; and integrate the u^(-1/4)
\[\frac{1}{20} \int\limits_{} u^{-1/4} du \rightarrow \frac{1}{20} \frac{u^{(-1+ 4)/4}}{(-1+4)/4}\]
u^(3/4)/(3/4) 4 * 4root(u^3) ------------ 20 * 3 4root(u^3) ------------ = 4root[5x^4-1]/15 5 * 3
forgot to ^3 my (5x^4 - 1)
\[\frac{\sqrt[4]{(5x^4 -1)^3}}{15} +C\]
You are great!!!!! So final answer: \[\frac{1}{20} (5x^{4} -1)^{3/4} + C?\]
that looks comparabley good :)
ack!!...1/15.....not 1/20
ahh!!
Fantastic!! lol - I almost messed that one up! Thank you so much! I have another problem I might post a new question so I can award medals - lol
im sure we will be around :)

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